Given the elementary symmetric polynomials $e_k(X_1,X_2,…,X_N)$ generated via
$$
\prod_{k=1}^{N} (t+X_k) = e_0t^N + e_1t^{N-1} + \cdots + e_N.
$$
How can one get the monomial symmetric functions $m_\lambda(X_1,X_2,…,X_N)$ as products and sums in $e_k$?
For example: $N=4$
$$
m_{(2,1,1,0)}=X_1^2X_2X_3 + \text{all permutations}= e_3\cdot e_1 – 4 e_4 ,
$$
$$
m_{(2,2,0,0)}=X_1^2X_2^2 + … = e_2^2-6e_4 – 2m_{(2,1,1,0)}=e_2^2 – 2e_3\cdot e_1 +2e_4
$$
It seems clear that the products on the RHS run over all partitions $\mu$ of $N$. For each $\lambda$ there should be a set of integers $C_{\lambda\mu}$ such that
$$
m_\lambda = \sum_\mu C_{\lambda \mu} \prod_j e_{\mu_j}
$$
is true. Putting all values of $c_{\lambda \mu}$ together, you get the following equation:
$$
\left(
\begin{matrix}
m_{4,0,0,0}\\
m_{3,1,0,0}\\
m_{2,2,0,0}\\
m_{2,1,1,0}\\
m_{1,1,1,1}\\
\end{matrix}
\right)=
\left(
\begin{matrix}
-4&+4&+2&-4&+1\\
+4&-1&-2&+1&0\\
+2&-2&+1&0&0\\
-4&+1&0&0&0\\
+1&0&0&0&0\\
\end{matrix}
\right)\left(
\begin{matrix}
e_{4}\\
e_3e_1\\
e_2^2\\
e_2e_1^2\\
e_1^4\\
\end{matrix}
\right)
.
$$
The transition matrix $C_4$ is symmetric. This also holds for $N=3$, where $C_3$ is
$$
\left(
\begin{matrix}
+3&-3&+1\\
-3&+1&0\\
+1&0&0\\
\end{matrix}
\right)
$$
and for $N=2$ we get
$$
\left(
\begin{matrix}
-2&+1\\
+1&0\\
\end{matrix}
\right).
$$
The main question is, if there exists a general formula for the entries of the matrices?
For a given $N$, is it a kind of composition of matrices $C_{k<N}$ and some few other matrices?
Beside that, the following is also of interest:
The matrices so far are symmetric and their entries sum up to $0$. Is this true in general? Is this related to (conjugate) Young Tableaux?
Balls&Boxes
If we bring the matrix on the LHS we get:
$$
\left(
\begin{matrix}
0 &0&0&0&1\\
0&0&0&1&4\\
0&0&1&2&6\\
0&1&2&5&12\\
1&4&6&12&24\\
\end{matrix}
\right)
\left(
\begin{matrix}
m_{4,0,0,0}\\
m_{3,1,0,0}\\
m_{2,2,0,0}\\
m_{2,1,1,0}\\
m_{1,1,1,1}\\
\end{matrix}
\right)=
\left(
\begin{matrix}
e_{4}\\
e_3e_1\\
e_2^2\\
e_2e_1^2\\
e_1^4\\
\end{matrix}
\right)\tag{*}
.
$$
Maybe it's easier to interpret these values, since they are all positive, in terms of balls, that have to be put into boxes, according to the following rules :
You are given $N$ balls. Your balls are now divided into parts (and bless god that, this is a math forum 🙂 according to a partition $\mu$. These are the products of $e_k$'s.
Now you are asked to put the balls partition-by-partition into a $N$ boxes. It is not allowed to put more than 1 ball in a box for the current partition.
The goal is to achieve a certain distribution of balls among the boxes, given by $\lambda$. These are the $m_\lambda$.
Power Sums
Or would it help to express $e_k$ as power sums via Newton-Girard formulas?
Here is the worked out example:
$$
\left(
\begin{matrix}
m_{4,0,0,0}\\
m_{3,1,0,0}\\
m_{2,2,0,0}\\
m_{2,1,1,0}\\
m_{1,1,1,1}\\
\end{matrix}
\right)=
\left(
\begin{matrix}
0& 0& 0& 0& 1\\
0& 0& 0& 1 &-1\\
0& 0& 1/2& 0& -1/2\\
0& 1/2& -1/2& -1& 1\\
1/24& -1/4 &+1/8 &1/3 &-1/4\\
\end{matrix}
\right)
\left(
\begin{matrix}
p_1^4\\
p_2p_1^2\\
p_2^2\\
p_3p_1\\
p_4^1\\
\end{matrix}
\right)
$$
Best Answer
Probably a simple general formula for your matrix $C_n=(C_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$, where $\mathcal P_n$ denotes the partitions of $n$ (ordered in decreasing lexicographic ordering) does not exist. However a number of things can be said, notably your above guesses can be confirmed. One thing that your examples suggest but which is false is that the matrix is "upper-left triangular", which fails from $n=6$ on; the reason is that the true triangularity is given by $ C_{\lambda,\mu}\neq0\implies \lambda^{tr}\leq\mu$ in the dominance order where $\lambda^{tr}$ is the transpose (conjugate) partition of $\lambda$, and lexicographic order for $n\geq6$ does not assign complementary positions to $\lambda$ and $\lambda^{tr}$ (nor does any ordering for $n$ sufficiently large).
As you guessed it is easier to study the inverse matrix $C_n^{-1}=(M_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$, whose entry $M_{\lambda,\mu}$ gives the coefficient of $m_\mu$ in $e_\lambda$. This nonnegative number equals number of $0$-$1$ matrices with row sums $\lambda_1,\lambda_2,\ldots$ and column sums $\mu_1,\mu_2,\ldots$, and in particular $M$ is a symmetric matrix (Stanley EC2, Proposition 7.4.1, Corollary 7.4.2). The argument for this combinatorial interpretation is basically the balls-into-boxes you suggested: each term in $e_\lambda$ comes from choosing a monomial in each factor $e_{\lambda_i}$, and this choice can be represented by putting in row $i$ a $1$ (ball) in the selected columns and zeros elsewhere; the product monomial is found by summing up the columns (and by symmetry only those monomials with weakly decreasing exponents (column sums) need to be considered). Since $M_n$ is symmetric and $C_n$ is its inverse, an easy argument shows that $C_n$ is also symmetric.
You suggested that the sum of all entries of $C_n$ is $0$. This fails for $n=0,1$, but is true for all larger values, which can be seen as follows. By the definition of $C_n$, if one takes its column sums (equivalently, if one left-multiplies by $(1~1\cdots1)$), then one gets the coefficients the $e_\mu$ in $h_n=\sum_{\lambda\in\mathcal P_n}m_\lambda$, the $n$-th complete homogeneous symmetric function (sum of all distinct monomials of degree $n$). One would like to know the sum of those coefficients. Now the relation between elementary and complete homogeneous symmetric functions is given by the generating series identity $$ (1-e_1X+e_2X^2-e_3X^3+\cdots)(1+h_1X+h_2X^2+\cdots)=1 $$ or equivalently $\sum_{i=0}^n(-1)^ne_ih_{n-i}=0^n$. You can prove that the mentioned sum is $0$ for $n\geq2$ by induction from the latter equation. A more conceptual way is to use the generating series identity and the algebraic independence of the $e_i$ for $i\geq1$, which means there is a ring morphism sending all of them to $1$, and the obvious fact that $$ (1-X+X^2-X^3+\cdots)^{-1}=1+X $$ This shows that the mentioned morphism sends $h_0$ and $h_1$ to $1$ and all other $h_i$ to $0$; this value is precisely the sum of coefficients of all $e_\lambda$ we were after.
Finally for the computation of the matrix $C_n$, it also seems best to view it as the inverse of the matrix $(M_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$, which becomes unitriangular after permuting its columns according to the transposition of partitions. However $M_{\lambda,\mu}$ is best computed not by counting $0$-$1$ matrices (their numbers grow up to $n!$ in the bottom-right corner), but rather via $$ M_{\lambda,\mu}= \sum_{\lambda\leq\nu\leq\mu^{tr}}K_{\nu,\lambda,}K_{\nu^{tr},\mu} $$ where $K_{\nu,\lambda}$ designates a Kostka number, the number of semi-standard Young tableaux of shape $\nu$ and weight (content) $\lambda$. This identity is proved bijectively by the asymmetric RSK-correspondence, a bijective correspondence between $0$-$1$ matrices and pairs of semi-standard tableaux of transpose shapes, their weights being given respectively by the column sums and the row sums of the matrix. The Kostka numbers involved are generally quite a bit smaller than $M_{\lambda,\mu}$, and moreover there are ways to compute them without counting; one method is to interpret them as weight multiplicities for $GL_n$ and use a formula like Freudenthal's recurrence for them. (The LiE program which I maintain does so, and will do this computation easily; one can get results online up to $n=9$ from this site, if one takes into account the fact that a partition is represented by the sequence of differences of successive parts: $[4,2,2,1,0,0,0,0,0]$ is represented as $[2,0,1,1,0,0,0,0]$.)
One could either compute $M_{\lambda,\mu}$ this way and invert the result, or invert the matrix of Kostka numbers and deduce from above formula, which can be interpreted as a matrix product, the formula $$ C_{\lambda,\mu}= \sum_{\lambda^{tr}\leq\nu\leq\mu}K'_{\lambda,\nu^{tr}}K'_{\mu,\nu} $$ where $(K'_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$ is the inverse of the Kostka matrix $(K_{\lambda,\mu})_{\lambda,\mu\in\mathcal P_n}$. I don't know very much about these "inverse Kostka numbers", but you will find some information about them in the answers to this MO question; I'm not sure this allows them to be computed more efficiently than by inverting the Kostka matrix.